state the law of conservation of momentum apply this law to explain the recoil of a gun when a shell is fired from it ?
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Statement: In absence of external force total momentum of system remains constant.
• • RECOIL OF GUN • •
Mass of shell = m1
Mass of gun = m2
Velocity of shell = v1
Recoil velocity of gun = v2
Initial momentum = Final momentum
0 = m1v1 + m2v2
v2 = -(m1v1) / m2
Recoil velocity of gun is (m1v1 / m2) in opposite direction of velocity of shell, hence -ve sign.
• • RECOIL OF GUN • •
Mass of shell = m1
Mass of gun = m2
Velocity of shell = v1
Recoil velocity of gun = v2
Initial momentum = Final momentum
0 = m1v1 + m2v2
v2 = -(m1v1) / m2
Recoil velocity of gun is (m1v1 / m2) in opposite direction of velocity of shell, hence -ve sign.
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Answer:
Mass of shell = m1
Mass of gun = m2
Velocity of shell = v1
Recoil velocity of gun = v2
Initial momentum = Final momentum
0 = m1v1 + m2v2
v2 = -(m1v1) / m2
Recoil velocity of gun is (m1v1 / m2) in opposite direction of velocity of shell, hence -ve sign.
Explanation:
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